On the zero of $f(x)=\int_0^x \ln(\sinh(z))dz$

Question

$$f(x)=\int_0^x \ln(\sinh(z))dz$$ I have observed that the only zero, besides $x=0$, of $f(x)$ is $x=x_0\approx2.146$. Is it possible that $x_0$ has a closed form in terms of $\pi$ or some other constant? If not, is there a possible series expansion for it? Much thanks and any help is appreciated.

Answer 1

This is a limited scope answer to address your subsidiary questions, since no other answers have been forthcoming so far:

Infinite Series for $\log(\sinh(z))$

$$\log(\sinh(z))=\log (z)-\sum _{k=1}^{\infty } \frac{ (-1)^k \zeta (2 k)}{k \pi ^{2 k}}z^{2 k}$$

which can easily be integrated term-wise when convergent to give

$$\int_0^x\log(\sinh(z))\; dz=-x+ x\log (x)-\sum _{k=1}^{\infty } \frac{ (-1)^k \zeta (2 k)}{k (2k+1)\pi ^{2 k}}x^{2 k+1}$$

There also appears to be a hyperbolic fourier type infinite series (analogous to the cos series for $\log(\sin z)$) which I am not sure how to prove (I've raised a separate question here):

$$\log(\sinh(z))=-\frac{1}{2} (i \pi )-\log (2)-\sum _{k=1}^{\infty } \frac{\cosh (2 k z)}{k}$$

which integrates to $$\int_0^x\log(\sinh(z))\; dz=-\frac{x}{2} (i \pi )-x \log (2)-\sum _{k=1}^{\infty } \frac{\sinh (2 k x)}{2 k^2}$$

Using this second formula Mathematica gives at the zero $x_0\approx2.14631$:

$$\frac{\text{Li}_2\left(e^{-2 (2.14631)}\right)-\text{Li}_2\left(e^{2 (2.14631)}\right)}{2.14631}=4 \log (2)+2 i \pi$$

which gives a strong indication I think that you won't find a closed form for $x_0$.